UDC 519.1,519.2
The number of vertices of xed degree in the
preferential attachment model with choice
∗
Y. Malyshkin
∗ Tver
State University,
Sadovij per. 35, Tver, 170002, Russia
The preferential attachment models are widely used to describe
dierent web and social network (random) graphs. We concentrate on some
generalizations of these models. Namely, a random tree under consideration
is constructed in the following way. Let d(1),d(2),... be i.i.d. integer-valued
random variables. At each step n a new vertex is introduced. Then we select
d(n) vertices, chosen from the old vertices with probabilities proportional to their
degrees and conditionally independent given d(n), and connect new vertex to the
vertex from the sample with largest degree. In preferential attachment without
choice d(n)=1 for all n. We establish the upper class law of the iterated logarithm
for the number of vertices of degree k at step n. The proof employs results of
stochastic approximation theory along with analysis of specied martingales and
system of the equations describing an evolution of the model.
Abstract.
Keywords:
random trees, stochastic approximation, preferential attachment.
1.
Introduction
In the present work, we study how the addition of choice aects the
Mori's preferential attachment model. Let describe the max-choice Mori's
preferential attachment tree model. This model is a time-indexed inductively constructed sequence of trees, built in the following way. First,
x number β > −1 and distribution of a random variable d with values in N. These are the parameters of our model. Consider set of vertices V = {vi }∞
i=1 . Dene a sequence of random trees {Tn }, n ∈ N, by
the following inductive rule. Let T1 be the one-edge tree which consists of vertices v1 and v2 and an edge between them. Given Tn , we
construct Tn+1 by adding one vertex and drawing one edge in the following way. First, we add a vertex vn+2 to Tn . So, for the set V (Tn+1 )
of vertices of Tn+1 we get V (Tn+1 ) = {vi , i = 1, ..., n + 2}. Note that
∑
vi ∈V (Tn ) degTn vi = 2n, where degTn vi is the degree of vi in Tn . Second, we draw an edge between vn+2 and Yn ∈ V (Tn ), which we choose by
the rule describe below.
So, for the set E(Tn+1 ) of edges of Tn+1 we get
∪
E(Tn+1 ) = E(Tn ) {vn+2 , Yn }. The randomness of Tn+1 given Tn is due
to randomness of Yn . Order the set V (Tn ) by the vertices degrees in Tn .
n
n
n
n
In other words, V (Tn ) = {v(1)
, ..., v(n+1)
}, degTn v(i)
≤ degTn v(i+1)
. Let
n
n
ϑj (n) denote the position of vertex vj in ordered set {v(1) , ..., v(n+1) }, i.e.
n
vj = v(ϑ
and if degTn vi = degTn vj and i < j , then ϑi (n) < ϑj (n).
j (n))
Consider i.i.d. random variables {Uni }n∈N,i∈N , distributed uniformly on
[0, 1]. Dene random variables Xni , i ∈ N with values in V (Tn ) as follow.
n
degTn v(1)
+β
n
Let Xni = v(1)
if Uni ≤ (2+β)n+β
and Xni = v(j) , 1 < j ≤ n + 1, if
∑j−1
n
degTn v(l)
+β
(2+β)n+β
∑j
n
+β
degTn v(l)
(2+β)n+β .
Let d1 , d2 , ... be i.i.d. random variable distributed as d. Finally, we take Yn as the vertex among
Xn1 ,...,Xndn with the largest degree. In the case of a tie, choose the vertex
with the largest index.
Remark 1 Since Tn+1 is well dened by Y1 , ..., Yn , all its parameters are
Fn+1 measurable.
Remark 2 Prove of the main result would require analysis of event An =
{(U1n , ..., Udnn ) ∈ Ddn } for random sets Di ∈ Ri , i ∈ N. Note that if |Di |
(here |B| stands for the Lebesgue measure of B ) does not depend on Fn ,
then An does not depend on Fn .
Let formulate our theorem. Let Nk (n) be the number of vertices of
degree k in tree Tn , Zk (n) = Nkn(n) and Wk (n) = (Z1 (n), ..., Zk (n)). Theorem 5.1 of [1] states that there is a point ρ∗k = (x∗1 , ..., x∗k ) and a positively
dened symmetric matrix Bk = (bi,j )1≤i,j≤k (both depends on β and distribution of d) such that n1/2 (Wk (n) − ρ∗k ) converge in distribution to
normal distribution N (0, B) as n → ∞. We will prove an upper class law
of the iterated logarithm for variables Nk (n).
2
Theorem 3 Let Ed < ∞. Then, for any k ∈ N one has
l=1
< Uni ≤
l=1
N (n) − nx∗ k
k lim sup √
≤1
2bk,k n ln ln n n→∞ 2.
a.s.
Proof of the main result
For x1 , ..., xk ∈ R+ , k ∈ N, dene functions
(
)m
1+β
h0 = 1, h1 (x1 ) =
P(d = m) x1
,
2+β
m=1
∞
∑
m 
m 
k−1
∑ (j + β)
(j
+
β)
 −
 ,
xj
hk (x1 , ..., xk ) =
P(d = m) 
xj
2
+
β
2
+
β
m=1
j=1
j=1
∞
∑

k
∑
fk (x1 , ..., xk ) = hk−1 (x1 , ..., xk−1 ) − hk (x1 , ..., xk ),
gk (x1 , ..., xk ) = f (x1 , ..., xk ) − xk .
As it is shown in [1],
b1,1
(
)m
∗1 + β
=1−
P(d = m) x1
,
2+β
m=1
∞
∑
P(N1 (n + 1) − N1 (n) = 1|Fn ) =
∞
∑
(
(
P(d = m) 1 −
m=1
and
N1 (n)(1 + β)
(2 + β)n + β
)m )
bk,k = hk (x∗1 , ..., x∗k ) + hk−1 (x∗1 , ..., x∗k−1 ),
gk (x∗1 , ..., x∗k ) = 0,
(
)
P (Nk (n + 1) − Nk (n) = 1|Fn ) = hk−1 Ze1 (n), ..., Zek (n) ,
(
)
P (Nk (n + 1) − Nk (n) = −1|Fn ) = hk Ze1 (n), ..., Zek (n)
(1)
(2)
(3)
for k > 1, where
1
Ni (n)
= Zi (n)
,
Zei (n) =
β
β
n + 2+β
1 + n(2+β)
i = 1, ..., k.
Also there are random Fn - measurable sets Dn+ (k, m), Dn− (k, m) ⊂ [0, 1]m ,
m, k, n ∈ N, such that
{Nk (n + 1) − Nk (n) = 1} = {(Un1 , ..., Undn ) ∈ Dn+ (k, dn )},
{Nk (n + 1) − Nk (n) = −1} = {(Un1 , ..., Undn ) ∈ Dn− (k, dn )}.
Therefore,
Nk (n + 1) − Nk (n) = 1{(Un1 , ..., Undn ) ∈ Dn+ (k, dn )}
−1{(Un1 , ..., Undn ) ∈ Dn− (k, dn )}.
Hence, from (2), (3) and denition of hk , we have for k > 1
m 
m 
k−1
∑
(j
+
β)
(j
+
β)
 −
 ,
Zej (n)
|Dn− (k, m)| = 
Zej (n)
2
+
β
2
+
β
j=1
j=1

k
∑

|Dn+ (k, m)| = 
k−1
∑
m

k−2
∑
m 
(j + β) 
(j + β)  
Zej (n)
−
Zej (n)
2
+
β
2+β
j=1
j=1
(
and for k = 1 we get |Dn− (1, m)| = 0 and |Dn+ (1, m)| = 1 −
(
(
−
+
Let α1,m
= 0, α1,m
= 1 − x∗1 (1+β)
2+β

+
αk,m
N1 (n)(1+β)
(2+β)n+β
)m )
and for k > 1
m 
m 
k−1
∑ (j + β)
(j
+
β)
 −
 ,
x∗j
= 
x∗j
2
+
β
2+β
j=1
j=1

−
αk,m
)m )
(
k
∑
m 
m 
k−2
∑ (j + β)
(j
+
β)
 −
 .
= 
x∗j
x∗j
2
+
β
2
+
β
j=1
j=1
k−1
∑
Introduce random Fn -measurable sets D∗+ (k, m, n) and D∗− (k, m, n) such
+
−
that |D∗+ (k, m, n)| = αk,m
, |D∗− (k, m, n)| = αk,m
. Moreover if |Dn+ (k, m)| >
+
+
αk,m
then D∗+ (k, m, n) ⊂ Dn+ (k, m), if |Dn+ (k, m)| ≤ αk,m
, then Dn+ (k, m) ⊂
−
+
−
−
D∗ (k, m, n), if |Dn (k, m)| > αk,m , then D∗ (k, m, n) ⊂ Dn− (k, m) and if
−
|Dn− (k, m)| ≤ αk,m
, then Dn− (k, m) ⊂ D∗− (k, m, n). Let
Xk (n) = 1{(Un1 , ..., Undn ) ∈ D∗+ (k, dn , n)}
−1{(Un1 , ..., Undn ) ∈ D∗− (k, dn , n)} − x∗k .
Note that due to Remark 2 and formula (1) we have that Xk (n), n ∈ N
are i.i.d. random variables and
EXk (n) = gk (x∗1 , ..., x∗k ) = 0.
Also EXk (n)2 = bk,k . Hence, by the law of the iterated logarithm for i.i.d.
random variables
∑
n
X
(i)
k
lim sup √ i=1
= 1 a.s.
2bk,k n ln ln n n→∞ Introduce random variables
ϵk (n) = 1{(Un1 , ..., Undn ) ∈ Dn+ (k, dn )\D∗+ (k, dn , n)}
−1{(Un1 , ..., Undn ) ∈ D∗+ (k, dn , n)\Dn+ (k, dn )}
−1{(Un1 , ..., Undn ) ∈ Dn− (k, dn )\D∗− (k, dn , n)}
Therefore,
+1{(Un1 , ..., Undn ) ∈ D∗− (k, dn , n)\Dn− (k, dn )}.
Nk (n + 1) − Nk (n) − x∗k = Xk (n) + ϵk (n).
.
To complete the proof we need to estimate
of the proof. Dene
∑n
. We give the sketch
i=1 ϵk (i)
Si (n) = Ni (n + 1) − Ni (n) − x∗i , Yk (n) = (S1 (n), ..., Sk (n)),
Fk (x1 , ..., xk ) = (f1 (x1 ), ..., fk (x1 , ..., xk )),
fk (n) = (Ze1 (n), ..., Zek (n)).
Ak (n) = (ϵ1 (n), ..., ϵk (n)), W
We get
E(Yk (n)|Fn ) = Fk (Ze1 (n), ..., Zek (n)) − Fk (x∗1 , ..., x∗k ).
Consequently,
(
)
fk (n) − ρ∗k + O(||W
fk (n) − ρ∗k ||2 )
E(Ak (n)|Fn ) = E(Yk (n)|Fn ) = ▽Fk W
(
= ▽Fk
1
n
(
(N1 (0), ..., Nk (0)) +
n
∑
))
fk (n) − ρ∗ ||2 ).
+ O(||W
k
Yk (i)
i=1
Since all eigenvalues of ▽Fk are negative, Ak (n) could be decomposed as
1
2
Ak (n) = A1k (n) + A2k (n) (with ϵ∑
i (n) = ϵi (n) + ϵi (n), i = 1, ..., k ) where
n
1
a.s. Ak (n) is of opposite sign to i=1 Yk (i) and E(A2k (n)|Fn ) = 0. Therefore,
logarithm for martingales (see, e.g., [2])
∑n by2 the law of the iterated
1/2
ln ln n)1/2 . Thus,
i=1 Ak (i) is of order (n
|
n
∑
i=1
≤|
n
∑
Sk (i)| = |
n
∑
Sk (i) + ϵ1k (n) + ϵ2k (n) + Xk (n)|
i=1
Sk (i) + ϵ1k (n)| + |ϵ2k (n)| + |Xk (n)| ≤ |
i=1
n
∑
Sk (i)| + |ϵ2k (n)| + |Xk (n)|
i=1
≤ ... ≤ |
n
∑
ϵ2k (i)| + |
i=1
n
∑
Xk (i)|.
i=1
We come to the desire statement.
References
1. Malyshkin Y. Preferential attachment combined with random number
of choices. ArXiv 1612.02229.
2. Stout W. A martingale analoque of Kolmogorov's law of iterated logarithm. Z. Vahr. Verw. Geb., 15, 279-290 (1970).